下面是工作代码与原型:

String.prototype.replaceAll = function(find, replace) {
    var str = this;
    return str.replace(new RegExp(find.replace(/([.*+?^=!:${}()|\[\]\/\\])/g, "\\$1"), 'g'), replace);
};

在最受欢迎的浏览器的最新版本中,您可以使用 ReplaceAll 如下:

let result = "1 abc 2 abc 3".replaceAll("abc", "xyz");
// `result` is "1 xyz 2 xyz 3"

但是,请检查我可以使用或其他兼容性表,首先确保您针对的浏览器首先添加了支持。

对于 Node.js 和旧/非当前浏览器的兼容性:

注意:不要在性能关键代码中使用下列解决方案。

作为一个简单的字体字符串的常规表达的替代品,您可以使用

str = "Test abc test test abc test...".split("abc").join("");

一般模式是

str.split(search).join(replacement)

这通常比在某些情况下更快,而不是使用替代所有和一个常见的表达式,但这似乎不再是现代浏览器的情况。

基准: https://jsben.ch/TZYzj

结论:

如果你有一个性能关键的使用案例(例如,处理数百条线),使用常规表达方法,但对于大多数典型的使用案例,这很好不需要担心特殊的字符。

这可以用常规表达式和旗帜g来解决,这意味着在找到第一场比赛后不停止。

function replaceAll(string, pattern, replacement) {
    return string.replace(new RegExp(pattern, "g"), replacement);
}

// or if you want myString.replaceAll("abc", "");

String.prototype.replaceAll = function(pattern, replacement) {
    return this.replace(new RegExp(pattern, "g"), replacement);
};

就像上面的分裂/合并解决方案一样,下面的解决方案与逃避字符没有任何问题,与常规表达方法不同。

function replaceAll(s, find, repl, caseOff, byChar) {
    if (arguments.length<2)
        return false;
    var destDel = ! repl;       // If destDel delete all keys from target
    var isString = !! byChar;   // If byChar, replace set of characters
    if (typeof find !== typeof repl && ! destDel)
        return false;
    if (isString && (typeof find !== "string"))
        return false;

    if (! isString && (typeof find === "string")) {
        return s.split(find).join(destDel ? "" : repl);
    }

    if ((! isString) && (! Array.isArray(find) ||
        (! Array.isArray(repl) && ! destDel)))
        return false;

    // If destOne replace all strings/characters by just one element
    var destOne = destDel ? false : (repl.length === 1);

    // Generally source and destination should have the same size
    if (! destOne && ! destDel && find.length !== repl.length)
        return false

    var prox, sUp, findUp, i, done;
    if (caseOff)  { // Case insensitive

    // Working with uppercase keys and target
    sUp = s.toUpperCase();
    if (isString)
       findUp = find.toUpperCase()
    else
       findUp = find.map(function(el) {
                    return el.toUpperCase();
                });
    }
    else { // Case sensitive
        sUp = s;
        findUp = find.slice(); // Clone array/string
    }

    done = new Array(find.length); // Size: number of keys
    done.fill(null);

    var pos = 0;  // Initial position in target s
    var r = "";   // Initial result
    var aux, winner;
    while (pos < s.length) {       // Scanning the target
        prox  = Number.MAX_SAFE_INTEGER;
        winner = -1;  // No winner at the start
        for (i=0; i<findUp.length; i++) // Find next occurence for each string
            if (done[i]!==-1) { // Key still alive

                // Never search for the word/char or is over?
                if (done[i] === null || done[i] < pos) {
                    aux = sUp.indexOf(findUp[i], pos);
                    done[i] = aux;  // Save the next occurrence
                }
                else
                    aux = done[i]   // Restore the position of last search

                if (aux < prox && aux !== -1) { // If next occurrence is minimum
                    winner = i; // Save it
                    prox = aux;
                }
        } // Not done

        if (winner === -1) { // No matches forward
            r += s.slice(pos);
            break;
        } // No winner

        // Found the character or string key in the target

        i = winner;  // Restore the winner
        r += s.slice(pos, prox); // Update piece before the match

        // Append the replacement in target
        if (! destDel)
            r += repl[destOne ? 0 : i];
        pos = prox + (isString ? 1 : findUp[i].length); // Go after match
    }  // Loop

    return r; // Return the resulting string
}

文档如下:

替代All Syntax ====== 替代All(s, find, [repl, caseOff, byChar) 参数 ==========“s” 是替代序列的目标. “find” 可以是序列或序列的序列. “repl” 应该是相同的类型“find” 或空的 如果“find” 是序列,它是一个简单的替代所有“find” 事件在“s” 由序列“repl” 如果“find” 是序列,它将取代

function l() {
    return console.log.apply(null, arguments);
}

var k = 0;
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      ["ri", "nea"], ["do", "fa"]));  // 1
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      ["ri", "nea"], ["do"])); // 2
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      ["ri", "nea"])); // 3
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
     "aeiou", "", "", true)); // 4
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      "aeiou", "a", "", true)); // 5
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      "aeiou", "uoiea", "", true)); // 6
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      "aeiou", "uoi", "", true)); // 7
l(++k, replaceAll("banana is a ripe fruit harvested near the river",
      ["ri", "nea"], ["do", "fa", "leg"])); // 8
l(++k, replaceAll("BANANA IS A RIPE FRUIT HARVESTED NEAR THE RIVER",
      ["ri", "nea"], ["do", "fa"])); // 9
l(++k, replaceAll("BANANA IS A RIPE FRUIT HARVESTED NEAR THE RIVER",
      ["ri", "nea"], ["do", "fa"], true)); // 10
return;

'a cat is not a caterpillar'.replace(/\bcat\b/gi,'dog');
//"a dog is not a caterpillar"

這是一個簡單的雷格斯,避免在大多數情況下取代字的部分. 然而,一個<unk> - 仍然被認為是字的邊界. 所以條件可以用在這種情況下,以避免取代線,如冷貓:

'a cat is not a cool-cat'.replace(/\bcat\b/gi,'dog');//wrong
//"a dog is not a cool-dog" -- nips
'a cat is not a cool-cat'.replace(/(?:\b([^-]))cat(?:\b([^-]))/gi,'$1dog$2');
//"a dog is not a cool-cat"

Regexp 不是唯一的替代多种现象的方法,远离它,思考灵活,思考分裂!

var newText = "the cat looks like a cat".split('cat').join('dog');

否则,要防止替代词部分 - 批准的答案也会做什么! 你可以通过常规的表达方式来围绕这个问题,我承认,有点复杂,并且作为一个惊喜,一个缓慢的,也:

var regText = "the cat looks like a cat".replace(/(?:(^|[^a-z]))(([^a-z]*)(?=cat)cat)(?![a-z])/gi,"$1dog");

结果与接受的答案相同,但是,在这个行上使用 /cat/g 表达式:

var oops = 'the cat looks like a cat, not a caterpillar or coolcat'.replace(/cat/g,'dog');
//returns "the dog looks like a dog, not a dogerpillar or cooldog" ??

var caterpillar = 'the cat looks like a cat, not a caterpillar or coolcat'.replace(/(?:(^|[^a-z]))(([^a-z]*)(?=cat)cat)(?![a-z])/gi,"$1dog");
//return "the dog looks like a dog, not a caterpillar or coolcat"

RegExp(常规表达式)对象 Regular-Expressions.info

在这种情况下,它显著简化表达,并提供更多的灵活性,如用正确的资本化替换或在一个行中替换两只猫和猫:

'Two cats are not 1 Cat! They\'re just cool-cats, you caterpillar'
   .replace(/(^|.\b)(cat)(s?\b.|$)/gi,function(all,char1,cat,char2)
    {
       // Check 1st, capitalize if required
       var replacement = (cat.charAt(0) === 'C' ? 'D' : 'd') + 'og';
       if (char1 === ' ' && char2 === 's')
       { // Replace plurals, too
           cat = replacement + 's';
       }
       else
       { // Do not replace if dashes are matched
           cat = char1 === '-' || char2 === '-' ? cat : replacement;
       }
       return char1 + cat + char2;//return replacement string
    });
//returns:
//Two dogs are not 1 Dog! They're just cool-cats, you caterpillar

如果使用图书馆是您的选择,那么您将获得与图书馆功能一起进行的测试和社区支持的好处。

// Include a reference to the string.js library and call it (for example) S.
str = S(str).replaceAll('abc', '').s;